IMO 1996 · Problem 4

The positive integers \(a\) and \(b\) are such that the numbers \(15a + 16b\) and \(16a - 15b\) are both squares of positive integers. What is the least possible value that can be taken on by the smaller of these two squares?

Official text of the problem. Official problem paper (PDF)

Answer

\[ \boxed{231361=481^2}. \]

Solution

Write \[ x^2=15a+16b,\qquad y^2=16a-15b, \] where \(x,y\) are positive integers. Then \[ x^2-31y^2=481(b-a), \] so \[ x^2\equiv31y^2\pmod{481}. \] We will show that both \(x\) and \(y\) are divisible by \(481=13\cdot37\).

Let \(p\) be either of the primes \(13\) and \(37\). We have \[ x^2\equiv31y^2\pmod p, \qquad 31^2\equiv-1\pmod p, \] where the second congruence follows from \(31^2+1=2\cdot13\cdot37\).

Suppose \(p\nmid y\). Since \(p\nmid31\), the first congruence also gives \(p\nmid x\). Raising it to the power \((p-1)/2\) and applying Fermat’s little theorem yields \[ 1\equiv x^{p-1} \equiv31^{(p-1)/2}y^{p-1} \equiv(-1)^{(p-1)/4} =-1\pmod p. \] Here \((p-1)/4\) is odd for both primes: it equals \(3\) or \(9\). This is impossible. Thus \(p\mid y\), and the congruence \(x^2\equiv31y^2\pmod p\) then gives \(p\mid x\).

Consequently, both \(13\) and \(37\) divide each of \(x,y\), so \(481\mid x\) and \(481\mid y\). Their positivity therefore implies \[ \min(x^2,y^2)\ge481^2. \]

Finally, take \[ a=31\cdot481,\qquad b=481. \] These are positive integers, and \[ 15a+16b=(15\cdot31+16)\cdot481=481^2, \] \[ 16a-15b=(16\cdot31-15)\cdot481=481^2. \] Thus the lower bound is attained, proving the answer.

The solution was checked against John Scholes' IMO solution archive (Kalva).